Question: Let $f(x)=\begin{cases} \log(3x)&\text{for }0<x<3 \\\\ (4-x)\log(9)&\text{for }x\geq 3 \end{cases}$ Is $f$ continuous at $x=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
For $f$ to be continuous at $x=3$, we need $\lim_{x\to 3}f(x)$ and $f(3)$ to exist and be equal. Since $3\geq 3$, the rule that applies to $x=3$ is $(4-x)\log(9)$. So $f(3)=(4-3)\log(9)=\log(9)$. Now let's analyze $\lim_{x\to 3}f(x)$. Finding $\lim_{x\to 3^{ +}}f(x)$ For $x$ -values larger than $3$, the appropriate rule for $f(x)$ is $(4-x)\log(9)$. Since $(4-x)\log(9)$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 3^{ +}}f(x) \\\\ &=\lim_{x\to 3^{ +}}[(4-x)\log(9)] \gray{(4-x)\log(9)\text{ is the rule for }x>3} \\\\ &=(4-3)\log(9) \gray{(4-x)\log(9)\text{ is continuous at }x=3} \\\\ &=\log(9) \end{aligned}$ Finding $\lim_{x\to 3^{ -}}f(x)$ For $x$ -values smaller than $3$, the appropriate rule for $f(x)$ is $\log(3x)$. Since $\log(3x)$ is continuous for $0<x<3$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 3^{ -}}f(x) \\\\ &=\lim_{x\to 3^{ -}}[\log(3x)] \gray{\log(3x)\text{ is the rule for }x<3} \\\\ &=\log(3(3)) \gray{\log(3x)\text{ is continuous at }x=3} \\\\ &=\log(9) \end{aligned}$ Conclusion We found that: $\lim_{x\to 3^{ +}}f(x)=\lim_{x\to 3^{ -}}f(x)=f(3)=\log(9)$ Since the one-sided limits are both equal to $f(3)$, we can determine that the two-sided limit $\lim_{x\to 3}f(x)$ is also equal to $f(3)$, and $f$ is continuous at $x=3$.